Explanation Given: The integral ∫ R ∫ 3 x y d A = 164 9 ; where R is the region bounded by the lines: x − 2 y = 0 , x − 2 y = − 4 , x + y = 4 and x + y = 1 . Formula used: Integral of ∫ x n d x = 1 n + 1 x n + 1 . Proof: First find all the points of intersection by solving each pair of equation of lines: For the lines x − 2 y = 0 and x + y = 4 , solve the pair of linear equation and get ( x , y ) = ( 8 3 , 4 3 ) , For the lines x − 2 y = − 4 and x + y = 4 , solve the pair of linear equation and get ( x , y ) = ( 4 3 , 8 3 ) Similarly, for the lines x − 2 y = − 4 and x + y = 1 , solve the pair of linear equation and get ( x , y ) = ( − 2 3 , 5 3 ) and for the lines x − 2 y = 0 and x + y = 1 , solve the pair of linear equation and get ( x , y ) = ( 2 3 , 1 2 ) . So, the integral ∫ R ∫ 3 x y d A bounded by these lineswill be: ∫ R ∫ 3 x y d A = ∫ − 2 3 2 3 ∫ 1 − x

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Chapter 14.8, Problem 15E

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To prove: To verify that the integral ∫R∫3xy dA=1649; where R is the region bounded by the lines: x−2y=0, x−2y=−4, x+y=4 and x+y=1.

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Chapter 14.8, Problem 14E

Chapter 14.8, Problem 16E

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To prove: To verify that the integral ∫ R ∫ 3 x y d A = 164 9 ; where R is the region bounded by the lines: x − 2 y = 0 , x − 2 y = − 4 , x + y = 4 and x + y = 1 .

Solution Summary: The author explains the formula used to prove the integral displaystyleundersetRint

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To prove: To verify that the integral ∫R∫3xy dA=1649; where R is the region bounded by the lines: x−2y=0, x−2y=−4, x+y=4 and x+y=1.

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Chapter 14 Solutions